# A circular coil has moment of inertia 0.8 kg m^2 around any diameter and is carrying current to produce a magnetic moment of 20 Am^2.

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A circular coil has moment of inertia 0.8 kg m2 around any diameter and is carrying current to produce a magnetic moment of 20 Am2. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of 4 T is applied along the vertical, it starts rotating, around its horizontal diameter. The angular speed the coil acquires after rotating by 60° will be

by (47.5k points)

by (48 points)

It can be solved using conservation of energy

by (48 points)

Ui + Ki  =  Uf + Kf

-MBcos60°+0 = - MBcos60° + 1/2​​​​​(Iw2​​​​​​)

On solving the above equation, we get

​​​​​​W = (MB/A)1/2

= 10.                              /

$\sqrt(MB/I)$

by (23.6k points)

By energy conservation